Symfony2中的多個動態防火牆和CAS服務器

Question

我正在Symfony開發一個應用程序來管理多所學校。 該應用程序有多個數據庫，每個學校一個，以及多個CAS服務器。

如果我只管理一所學校，配置將如下：

# config.yml
be_simple_sso_auth:
    admin_sso:
        protocol:
            id: cas
            version: 2
        server:
            id: cas
            login_url: https://cas01.example.com/SCHOOLID/login
            logout_url: https://cas01.example.com/SCHOOL_ID/logout
            validation_url: https://cas01.example.com/SCHOOL_ID/serviceValidate

# security.yml
firewalls:
    school:
        pattern: ^/school/.*$
        trusted_sso:
            manager: admin_sso
            login_action: false 
            logout_action: false 
            create_users: true
            created_users_roles: [ROLE_USER, ROLE_ADMIN]
            login_path: /school/login
            check_path: /school/login_check
        logout:
            path:   /school/logout
            target: /school

有一所學校一切正常。

每所學校都通過app.com/school/ID路徑訪問該應用程序，例如app.com/school/29,app.com/school/54 ...

我想知道是否有辦法根據ID有多個動態防火牆。 並使用此ID重定向每個CAS URL：

https://cas01.example.com/school_29/login ， https://cas01.example.com/school_54/login ...

-----------更新13/12/12 -----------

我創建了一個新文件：app / config / cas.php，我添加了一些CAS服務器設置

# CAS 14
$container->loadFromExtension('be_simple_sso_auth', array(
    'cas_14' => array(
        'protocol' => array(
            'id' => 'cas',
            'version' => '2'
        ),
        'server' => array(
            'id' => 'cas',
            'login_url' => 'https://cas01.example.com/14/login',
            'logout_url' => 'https://cas01.example.com/14/logout',
            'validation_url' => 'https://cas01.example.com/14/serviceValidate',
        ),
    ),

));

# CAS 15
$container->loadFromExtension('be_simple_sso_auth', array(
    'cas_15' => array(
        'protocol' => array(
            'id' => 'cas',
            'version' => '2'
        ),
        'server' => array(
            'id' => 'cas',
            'login_url' => 'https://cas01.example.com/15/login',
            'logout_url' => 'https://cas01.example.com/15/logout',
            'validation_url' => 'https://cas01.example.com/15/serviceValidate',
        ),
    ),

));

我在config.yml中導入此文件

imports:
    - { resource: parameters.yml }
    - { resource: cas.php }
    - { resource: security.yml }

我為每所學校添加了一個新的防火牆：

firewalls:
    backend_14:
        pattern: ^/backend/school/14/.*$
        trusted_sso:
            manager: cas_14
            login_action: false #BeSimpleSsoAuthBundle:TrustedSso:login
            logout_action: false #BeSimpleSsoAuthBundle:TrustedSso:logout
            create_users: true
            created_users_roles: [ROLE_USER, ROLE_ADMIN]
            login_path: /backend/school/14/login
            check_path: /backend/school/14/login_check
        logout:
            path:   /backend/school/logout
            target: /backend

    backend_15:
        pattern: ^/backend/school/15/.*$
        trusted_sso:
            manager: cas_15
            login_action: false #BeSimpleSsoAuthBundle:TrustedSso:login
            logout_action: false #BeSimpleSsoAuthBundle:TrustedSso:logout
            create_users: true
            created_users_roles: [ROLE_USER, ROLE_ADMIN]
            login_path: /backend/school/15/login
            check_path: /backend/school/15/login_check
        logout:
            path:   /backend/school/logout
            target: /backend

一切順利！

現在我正在嘗試從Entity School生成所有cas.php配置動態。 首先，我嘗試在SchoolController中創建一個方法

public function loadCasConfig()
{
    $em = $this->getDoctrine()->getManager();

    $schools= $em->getRepository('SchoolBundle:School')
                  ->findBy(array(), array('name'=> 'ASC'));


    foreach ($schools as $school) {

        $cas_name = 'cas_'.$school->getId();

        $container->loadFromExtension('be_simple_sso_auth', array(
            "$cas_name" => array(
                'protocol' => array(
                    'id' => 'cas',
                    'version' => '2'
                ),
                'server' => array(
                    'id' => 'cas',
                    'login_url' => "https://cas01.example.com/$school->getId()/login",
                    'logout_url' => "https://cas01.example.com/$school->getId()/logout",
                    'validation_url' => "https://cas01.example.com/$school->getId()/serviceValidate",
                ),
            ),

        ));

    }
}

並在cas.php文件中調用它

<?php   

use Comp\BackendBundle\Controller\SchoolController;

SchoolController::loadCasConfig();

但我有這個例外：

FileLoaderLoadException: Cannot import resource     
"C:\wamp\www\comp\app/config\cas.php" from     
"C:\wamp\www\comp\app/config\config.yml". (Runtime Notice: Non-static method     
Comp\BackendBundle\Controller\SchoolController::loadCasConfig() should not be     
called statically, assuming $this from incompatible context in     C:\wamp\www\comp\app\config\cas.php line 5)

:(。然后我嘗試在cas.php文件中插入方法代碼：

use Doctrine\ORM\EntityManager;
use Comp\SchoolBundle\Entity\School;

$em = $this->getDoctrine()->getManager();

$schools= $em->getRepository('SchoolBundle:School')
              ->findBy(array(), array('name'=> 'ASC'));


foreach ($schools as $school) {

    $cas_name = 'cas_'.$school->getId();

    $container->loadFromExtension('be_simple_sso_auth', array(
        "$cas_name" => array(
            'protocol' => array(
                'id' => 'cas',
                'version' => '2'
            ),
            'server' => array(
                'id' => 'cas',
                'login_url' => "https://cas01.example.com/$school->getId()/login",
                'logout_url' => "https://cas01.example.com/$school->getId()/logout",
                'validation_url' => "https://cas01.example.com/$school->getId()/serviceValidate",
            ),
        ),

    ));

}

現在我有：

FatalErrorException: Error: Call to undefined method 
Symfony\Component\DependencyInjection\Loader\PhpFileLoader::getDoctrine() in 
C:\wamp\www\comp\app\config\cas.php line 11

我想知道如何動態生成文件cas.php，從數據庫中獲取數據。

Answer 1

代碼示例：

public function registerContainerConfiguration(LoaderInterface $loader)
{
    $loader->load(__DIR__.'/config/sites/' . $this->_activeSite . '/config.yml');
    $configPath = __DIR__ . '/config/sites/' . $this->_activeSite . '/config_' . $this->getEnvironment() . '.yml';
    if (file_exists($configPath)) {
        $loader->load($configPath);
    }

    $loader->load(__DIR__.'/config/servers/' . $this->getServer() . '.yml');
    $loader->load(__DIR__.'/config/config_'.$this->getEnvironment().'.yml');

    //per-site overrides
    if (file_exists(__DIR__.'/config/sites/local_' . $this->_activeSite . '.yml')) {
        $loader->load(__DIR__.'/config/sites/local_' . $this->_activeSite . '.yml');
    }
}

Answer 2

我們有類似的問題，當一個平台被多個網站使用時，所以我們有解決方法，現在每個網站都有自己的security.yml ，導入主security.yml

Answer 3

public function loadCasConfig（）不是靜態方法。 創建一個靜態...然后調用。

像這樣：

public function loadCasConfig()
{
    $em = $this->getDoctrine()->getManager();

    $schools= $em->getRepository('SchoolBundle:School')
                  ->findBy(array(), array('name'=> 'ASC'));


    foreach ($schools as $school) {

        $cas_name = 'cas_'.$school->getId();

        $container->loadFromExtension('be_simple_sso_auth', array(
            "$cas_name" => array(
                'protocol' => array(
                    'id' => 'cas',
                    'version' => '2'
                ),
                'server' => array(
                    'id' => 'cas',
                    'login_url' => "https://cas01.example.com/$school->getId()/login",
                    'logout_url' => "https://cas01.example.com/$school->getId()/logout",
                    'validation_url' => "https://cas01.example.com/$school->getId()/serviceValidate",
                ),
            ),

        ));

    }
}

public static function loadCasConfigStatic(){ 
   $my = new static; 
   return $my->loadCasConfig(); 
} 

然后（你的cas.php）：

<?php   

use Comp\BackendBundle\Controller\SchoolController;

SchoolController::loadCasConfigStatic();